The efficiency of a carnot cycle is `1//6` . On decreasing the tempertaure of the sink by `65^(@)C`, the efficiency increases to `1//3` .Calculate the tempretaure of source and sink .

Answer

`117^(@)C, 52^(@)`

Explanation

`(1)/(6)=1 - (T_(L))/(T_(H))`
`T_(H)=(6)/(5)T_(L)`
`(1)/(3)= 1-((T_(L) - 65))/(T_(H))`
`T_(L) = 325 K = 52 ^(@)C`